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I' trying to solve for the surface temperature Ts=((n+1)So(1-A)4)1/4, and thus So=L/4𝞹a2 but I cannot seem to find the value for solar flux or its luminosity.

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    $\begingroup$ Your equations seem a little messed up, and I don't trust a blind MathJax conversion. Are there some exponents in there that you left out? $\endgroup$ – Nathan Tuggy Mar 7 '18 at 3:09
  • $\begingroup$ @NathanTuggy I agree. It's probably something like one of these, but I don't know what the (n+1) is en.wikipedia.org/wiki/… See also cow: space.stackexchange.com/q/25671/12102 $\endgroup$ – uhoh Mar 7 '18 at 3:21
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The solar flux at 1 AU (Earth's distance from the sun) is ~1361 W/m2. Flux falls off as the square of distance. Saturn, on average -- and thus Titan, on average -- is about 9.58 AU from the sun, so the falloff factor is 1 / 9.582 or 0.0109.

Titan should average about 1361 x 0.0109 = 14.8 W/m2 when not shadowed by the gas giant.

When Saturn is at its furthest from the sun, 10.12 AU, the flux would be closer to 13.3 W/m2.

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  • $\begingroup$ @Haley can you consider clicking "Accept" (the check-mark icon) to the left of this answer? That way other readers in the future will see that the question has an accepted answer, and will be better able to find it. $\endgroup$ – uhoh Mar 7 '18 at 2:56

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